Title: PCI 6th Edition
1PCI 6th Edition
- Flexural Component Design
2Presentation Outline
- Whats new to ACI 318
- Gravity Loads
- Load Effects
- Concrete Stress Distribution
- Nominal Flexural Strength
- Flexural Strength Reduction Factors
- Shear Strength
- Torsion
- Serviceability Requirements
3New to ACI 318 02
- Load Combinations
- Stress limits
- Member Classification
- Strength Reduction factor is a function of
reinforcement strain - Minimum shear reinforcement requirements
- Torsion Design Method
4Load Combinations
- U 1.4 (D F)
- U 1.2 (D F T) 1.6 (L H) 0.5 (Lr or S
or R) - U 1.2D 1.6 (Lr or S or R) (1.0L or 0.8W)
- U 1.2D 1.6W 1.0L 0.5(Lr or S or R)
- U 1.2D 1.0E 1.0L 0.2S
- U 0.9D 1.6W 1.6H
- U 0.9D 1.0E 1.6H
5Comparison of Load Combinations
- U1.2D 1.6 L 2002
- U 1.4D 1.7L 1999
- If L.75D
- i.e. a 10 reduction in required strength
6Classifications
- No Bottom Tensile Stress Limits
- Classify Members Strength Reduction Factor
- Tension-Controlled
- Transition
- Compression Controlled
- Three Tensile Stress Classifications
- Class U Un-cracked
- Class T Transition
- Class C Cracked
7 - Copied from ACI 318 2002, ACI 318-02 table R18.3.3
8Class C Members
- Stress Analysis Based on Cracked Section
Properties - No Compression Stress limit
- No Tension Stress limit
- Increase awareness on serviceability
- Crack Control
- Displacements
- Side Skin Reinforcement
9Minimum Shear Reinforcing
1999
2002
10System Loads
- Gravity Load Systems
- Beams
- Columns
- Floor Member Double Tees, Hollow Core
- Spandrels
- Tributary Area
- Floor members, actual top area
- Beams and spandrels
- Load distribution
- Load path
- Floor members ? spandrels or beams ? Columns
11Live Load Reduction
- Live Loads can be reduced based on
-
- Where
- KLL 1
- Lo Unreduced live load and
- At tributary area
12Live Load Reduction
- Or the alternative floor reduction shall not
exceed - or
- Where
- R reduction 40
- r .08
13Member Shear and Moment
- Shear and moments on members can be found using
statics methods and beam tables from Chapter 11
14Strength Design
- Strength design is based using the rectangular
stress block - The stress in the prestressing steel at nominal
strength, fps, can be determined by strain
compatibility or by an approximate empirical
equation - For elements with compression reinforcement, the
nominal strength can be calculated by assuming
that the compression reinforcement yields. Then
verified. - The designer will normally choose a section and
reinforcement and then determine if it meets the
basic design strength requirement
15Concrete Stress Distribution
- Parabolic distribution
- Equivalent rectangular distribution
16Stress Block Theory
- Stress-Strain relationship
- is not constant
fc6,000 psi
fc3,000 psi
17Stress Block Theory
- Stress-Strain relationship
- Stress-strain can be modeled by
Where strain at max.
stress
and max stress
18Stress Block Theory
- The Whitney stress block is a simplified stress
distribution that shares the same centroid and
total force as the real stress distribution
19Equivalent Stress Block b1 Definition
- b1 0.85
- when fc lt 3,000 psi
- b1 0.65
- when fc gt 8,000 psi
20Design Strength
- Mild Reinforcement Non - Prestressed
- Prestress Reinforcement
21Strength Design Flowchart
- Figure 4.2.1.2 page 4-9
- Non-Prestressed Path
- Prestressed Path
22Non-Prestressed Members
- Find depth of compression block
23Depth of Compression Block
- Where
- As is the area of tension steel
- As is the area of compression steel
- fy is the mild steel yield strength
Assumes compression steel yields
24Flanged Sections
- Checked to verify that the compression block is
truly rectangular
25Compression Block Area
- If compression block is rectangular, the flanged
section can be designed as a rectangular beam
26Compression Block Area
- If the compression block is not rectangular (agt
hf),
To find a
27Determine Neutral Axis
- From statics and strain compatibility
28Check Compression Steel
- Verify that compression steel has reached yield
using strain compatibility
29Compression Comments
- By strain compatibility, compression steel yields
if - If compression steel has not yielded, calculation
for a must be revised by substituting actual
stress for yield stress - Non prestressed members should always be tension
controlled, therefore c / dt lt 0.375 - Add compression reinforcement to create tesnion
controlled secions
30Moment Capacity
- 2 equations
- rectangular stress block in the flange section
- rectangular stress block in flange and stem
section
31Strength Design Flowchart
Figure 4.2.1.2page 4-9 Non- Prestressed
Path Prestressed Path
32(No Transcript)
33Stress in Strand
- fse - stress in the strand after losses
- fpu - is the ultimate strength of the strand
- fps - stress in the strand at nominal strength
34Stress in Strand
- Typically the jacking force is 65 or greater
- The short term losses at midspan are about 10 or
less - The long term losses at midspan are about 20 or
less
35Stress in Strand
- Nearly all prestressed concrete is bonded
36Stress in Strand
- Prestressed Bonded reinforcement
- gp factor for type of prestressing strand, see
ACI 18.0 - .55 for fpy/fpu not less than .80
- .45 for fpy/fpu not less than .85
- .28 for fpy/fpu not less than .90 (Low
Relaxation Strand) - rp prestressing reinforcement ratio
37Determine Compression Block
38Compression Block Height
- Where
- Aps - area of prestressing steel
- fps - prestressing steel strength
39Flange Sections Check
40Compression Steel Check
- Verify that compression steel has reached yield
using strain compatibility
41Moment Capacity
- 2 Equations
- rectangular stress block in flange section
- rectangular stress block in flange and stem
section
42Flexural Strength Reduction Factor
- Based on primary reinforcement strain
- Strain is an indication of failure mechanism
- Three Regions
43Member Classification
44Compression Controlled
- e lt 0.002 at extreme steel tension fiber or
- c/dt gt 0.600
- 0.70 with spiral ties
- 0.65 with stirrups
45Tension Controlled
- e gt 0.005 at extreme steel tension fiber, or
- c/dt lt 0.375
- f 0.90 with spiral ties or stirrups
46Transition Zone
- 0.002 lt e lt 0.005 at extreme steel tension fiber,
or - 0.375 lt c/dt lt 0.6
- f 0.57 67(e) or
- f 0.48 83(e) with spiral ties
- f 0.37 0.20/(c/dt) or
- f 0.23 0.25/(c/dt) with stirrups
47Strand Slip Regions
- ACI Section 9.3.2.7
- where the strand embedment length is less than
the development length - f 0.75
48Limits of Reinforcement
- To prevent failure immediately upon cracking,
Minimum As is determined by - As,min is allowed to be waived if tensile
reinforcement is 1/3 greater than required by
analysis
49Limits of Reinforcement
- The flexural member must also have adequate
reinforcement to resist the cracking moment - Where
Correction for initial stresses on non-composite,
prior to topping placement
Section after composite has been applied,
including prestress forces
50Critical Sections
51Horizontal Shear
- ACI requires that the interface between the
composite and non-composite, be intentionally
roughened, clean and free of laitance - Experience and tests have shown that normal
methods used for finishing precast components
qualifies as intentionally roughened
52Horizontal Shear, Fh Positive Moment Region
- Based on the force transferred in topping (page
4-53)
53Horizontal Shear, Fh Negative Moment Region
- Based on the force transferred in topping (page
4-53)
54Unreinforced Horizontal Shear
- Where
- f 0.75
- bv width of shear area
- lvh - length of the member subject to shear,
1/2 the span for simply supported members
55Reinforced Horizontal Shear
- Where
- f 0.75
- rv - shear reinforcement ratio
- Acs - Area of shear reinforcement
- me - Effective shear friction coefficient
56Shear Friction Coefficient
57Shear Resistance by Non-Prestressed Concrete
- Shear strength for non-prestressed sections
58Prestress Concrete Shear Capacity
- Where
- ACI Eq 11-9
- Effective prestress must be 0.4fpu
- Accounts for shear combined with moment
- May be used unless more detail is required
59Prestress Concrete Shear Capacity
- Concrete shear strength is minimum is
- Maximum allowed shear resistance from concrete
is
60Shear Capacity, Prestressed
- Resistance by concrete when diagonal cracking is
a result of combined shear and moment
Where Vi and Mmax - factored externally
applied loads e.g. no self weight Vd - is
un-factored dead load shear
61Shear Capacity, Prestressed
- Resistance by concrete when diagonal cracking is
a result of principal tensile stress in the web
is in excess of cracking stress.
Where Vp the vertical component of
effective prestress force (harped or draped
strand only)
62Vcmax
- Shear capacity is the minimum of Vc, or if a
detailed analysis is used the minimum of Vci or
Vcw
63Shear Steel
64Shear Steel Minimum Requirements
- Non-prestressed members
- Prestressed members
Remember both legs of a stirrup count for Av
65Torsion
- Current ACI
- Based on compact sections
- Greater degree of fixity than PC can provide
- Provision for alternate solution
- Zia, Paul and Hsu, T.C., Design for Torsion and
Shear in Prestressed Concrete, Preprint 3424,
American Society of Civil Engineers, October,
1978. Reprinted in revised form in PCI JOURNAL,
V. 49, No. 3, May-June 2004.
66Torsion
- For members loaded two sides, such as inverted
tee beams, find the worst case condition with
full load on one side, and dead load on the other
67Torsion
- In order to neglect Torsion
- Where
- Tu(min) minimum torsional strength provided by
concrete
68Minimum Torsional Strength
- Where
- x and y - are short and long side, respectively
of a component rectangle - g - is the prestress factor
69Prestress Factor, g
- For Prestressed Members
- Where
- fpc level of prestress after losses
70Maximum Torsional Strength
- Avoid compression failures due to over
reinforcing - Where
71Maximum Shear Strength
- Avoid compression failures due to over reinforcing
72Torsion/Shear Relationship
- Determine the torsion carried by the concrete
- Where
- Tc and Vc - concrete resistance under pure
torsion and shear respectively - Tc and Vc - portions of the concrete
resistance of torsion and shear
73Torsion/Shear Relationship
- Determine the shear carried by the concrete
74Torsion Steel Design
- Provide stirrups for torsion moment - in addition
to shear - Where
- x and y - short and long dimensions of the
closed stirrup
75Torsion Steel Design
- Minimum area of closed stirrups is limited by
76Longitudinal Torsion Steel
- Provide longitudinal steel for torsion based on
equation -
- or
- Whichever greater
77Longitudinal Steel limits
- The factor in
- the second equation need not exceed
78Detailing Requirements, Stirrups
- 135 degree hooks are required unless sufficient
cover is supplied - The 135 degree stirrup hooks are to be anchored
around a longitudinal bar - Torsion steel is in addition to shear steel
79Detailing Requirements, Longitudinal Steel
- Placement of the bars should be around the
perimeter - Spacing should spaced at no more than 12 inches
- Longitudinal torsion steel must be in addition to
required flexural steel (note at ends flexural
demand reduces) - Prestressing strand is permitted (_at_ 60ksi)
- The critical section is at the end of simply
supported members, therefore U-bars may be
required to meet bar development requirements
80Serviceability Requirements
- Three classifications for prestressed components
- Class U Uncracked
- Class T Transition
- Class C Cracked
Stress
81Uncracked Section
- Table 4.2.2.1 (Page 4.24)
- Easiest computation
- Use traditional mechanics of materials methods to
determine stresses, gross section and deflection. - No crack control or side skin reinforcement
requirements
82Transition Section
- Table 4.2.2.1 (Page 4.24)
- Use traditional mechanics of materials methods to
determine stresses only. - Use bilinear cracked section to determine
deflection - No crack control or side skin reinforcement
requirements
83Cracked Section
- Table 4.2.2.1 (Page 4.24)
- Iterative process
- Use bilinear cracked section to determine
deflection and to determine member stresses - Must use crack control steel per ACI 10.6.4
modified by ACI 18.4.4.1 and ACI 10.6.7
84Cracked Section Stress Calculation
- Class C member require stress to be check using a
Cracked Transformed Section - The reinforcement spacing requirements must be
adhered to
85Cracked Transformed Section Property Calculation
Steps
- Step 1 Determine if section is cracked
- Step 2 Estimate Decompression Force in Strand
- Step 3 Estimate Decompression Force in mild
reinforcement (if any) - Step 4 Create an equivalent force in topping if
present - Step 5 Calculate transformed section of all
elements and modular ratios - Step 6 Iterate the location of the neutral axis
until the normal stress at this level is zero - Step 7 Check Results with a a moment and force
equilibrium set of equations
86Steel Stress
- fdc decompression stress
- stress in the strand when the surrounding
concrete stress is zero Conservative to use,
fse (stress after losses) when no additional mild
steel is present.
87Simple Example
88(No Transcript)
89(No Transcript)
90(No Transcript)
91(No Transcript)
92Deflection Calculation Bilinear Cracked Section
- Deflection before the member has cracked is
calculated using the gross (uncracked) moment of
inertia, Ig - Additional deflection after cracking is
calculated using the moment of inertia of the
cracked section Icr
93Effective Moment of Inertia
Where ftl final stress fl stress due
to live load fr modulus of rupture
94Prestress Losses
- Prestressing losses
- Sources of total prestress loss (TL)
- TL ES CR SH RE
- Elastic Shortening (SH)
- Creep (CR)
- Shrinkage (SH)
- Relaxation of tendons (RE)
95Elastic Shortening
- Caused by the prestressed force in the precast
member - Where
- Kes 1.0 for pre-tensioned members
- Eps modulus of elasticity of prestressing
tendons (about 28,500 ksi) - Eci modulus of elasticity of concrete at time
prestress is applied - fcir net compressive stress in concrete at
center of gravity of prestressing force
immediately after the prestress has been applied
to the concrete
96fcir
- Where
- Pi initial prestress force (after anchorage
seating loss) - e eccentricity of center of gravity of tendons
with respect to center of gravity of concrete at
the cross section considered - Mg bending moment due to dead weight of
prestressed member and any other permanent loads
in place at time of prestressing - Kcir 0.9 for pretensioned members
97Creep
- Creep (CR)
- Caused by stress in the concrete
- Where
- Kcr 2.0 normal weight concrete
- 1.6 sand-lightweight concrete
- fcds stress in concrete at center of gravity
of prestressing force due to all uperimposed
permanent dead loads that are applied to the
member after it has been prestressed
98fcds
- Where
- Msd moment due to all superimposed permanent
dead and sustained loads applied after
prestressing
99Shrinkage
- Volume change determined by section and
environment - Where
- Ksh 1.0 for pretensioned members
- V/S volume-to-surface ratio
- R.H. average ambient relative humidity from
map
100Relative Humidity
- Page 3-114 Figure 3.10.12
101Relaxation
- Relaxation of prestressing tendons is based on
the strand properties - Where
- Kre and J - Tabulated in the PCI handbook
- C - Tabulated or by empirical equations in the
PCI handbook
102Relaxation Table
- Values for Kre and J for given strand
- Table 4.7.3.1 page 4-85
103Relaxation Table Values for C
- fpi initial stress in prestress strand
- fpu ultimate stress for prestress strand
- Table 4.7.3.2 (Page 4-86)
104Prestress Transfer Length
- Transfer length Length when the stress in the
strand is applied to the concrete - Transfer length is not used to calculate capacity
105Prestress Development Length
- Development length - length required to develop
ultimate strand capacity - Development length is not used to calculate
stresses in the member
106Beam Ledge Geometry
107Beam Ledge Design
- For Concentrated loads where s gt bt hl, find
the lesser of
108Beam Ledge Design
- For Concentrated loads where s lt bt hl, find
the lesser of
109Beam Ledge Reinforcement
- For continuous loads or closely spaced
concentrated loads - Ledge reinforcement should be provided by 3
checks - As, cantilevered bending of ledge
- Al, longitudinal bending of ledge
- Ash, shear of ledge
110Beam Ledge Reinforcement
- Transverse (cantilever) bending reinforcement, As
- Uniformly spaced over width of 6hl on either side
of the bearing
- Not to exceed half the distance to the next load
- Bar spacing should not exceed the ledge depth,
hl, or 18 in
111Longitudinal Ledge Reinforcement
- Placed in both the top and bottom of the ledge
portion of the beam - Where
- dl - is the depth of steel
-
- U-bars or hooked bars may
- be required to develop
- reinforcement at the end
- of the ledge
112Hanger Reinforcement
- Required for attachment of the ledge to the web
- Distribution and spacing of Ash reinforcement
should follow the same guidelines as for As
113Hanger (Shear) Ledge Reinforcement
- Ash is not additive to shear and torsion
reinforcement - m is a modification factor which can be
derived, and is dependent on beam section
geometry. PCI 6th edition has design aids on
table 4.5.4.1
114Dap Design
- (1) Flexure (cantilever bending) and axial
tension in the extended end. Provide flexural
reinforcement, Af, plus axial tension
reinforcement, An.
115Dap Design
- (2) Direct shear at the junction of the dap and
the main body of the member. Provide shear
friction steel, composed of Avf Ah, plus axial
tension reinforcement, An
116Dap Design
- (3) Diagonal tension emanating from the
re-entrant corner. Provide shear reinforcement,
Ash
117Dap Design
- (4) Diagonal tension in the extended end. Provide
shear reinforcement composed of Ah and Av
118Dap Design
- (5) Diagonal tension in the undapped portion.
This is resisted by providing a full development
length for As beyond the potential crack.
119Dap Reinforcement
- 5 Main Areas of Steel
- Tension - As
- Shear steel - Ah
- Diagonal cracking Ash, Ash
- Dap Shear Steel - Av
120Tension Steel As
- The horizontal reinforcement is determined in a
manner similar to that for column corbels
121Shear Steel Ah
- The potential vertical crack (2) is resisted by a
combination of As and Ah
122Shear Steel Ah
- Note the development ld of Ah beyond the assumed
crack plane. Ah is usually a U-bar such that the
bar is developed in the dap
123Diagonal Cracking Steel Ash
- The reinforcement required to resist diagonal
tension cracking starting from the re-entrant
corner (3) can be calculated from
124Dap Shear Steel Av
- Additional reinforcement for Crack (4) is
required in the extended end, such that
125Dap Shear Steel Av
- At least one-half of the reinforcement required
in this area should be placed vertically. Thus
126Dap Limitations and Considerations
- Design Condition as a dap if any of the following
apply - The depth of the recess exceeds 0.2H or 8 in.
- The width of the recess (lp) exceeds 12 in.
- For members less than 8 in. wide, less than
one-half of the main flexural reinforcement
extends to the end of the member above the dap - For members 8 in. or more wide, less than
one-third of the main flexural reinforcement
extends to the end of the member above the dap
127Questions?